Filters on a set of filters, are they equivalent to just filters?
$begingroup$
Let $F(X)$ be the set of all filters (including the improper filter) on a poset $X$, ordered reversely to set-theoretic inclusion of filters.
Let $U$ be a set. Is $F(F(mathscr{P}U))$ order isomorphic to $F(mathscr{P}U)$?
set-theory filters
asked Jul 19 '14 at 18:03
portonporton
1,92111229
$endgroup$
add a comment |
$begingroup$
Let $F(X)$ be the set of all filters (including the improper filter) on a poset $X$, ordered reversely to set-theoretic inclusion of filters.
Let $U$ be a set. Is $F(F(mathscr{P}U))$ order isomorphic to $F(mathscr{P}U)$?
set-theory filters
asked Jul 19 '14 at 18:03
portonporton
1,92111229
$endgroup$
$begingroup$
Have you checked the finite case where $U$ has a trivial poset structure (i.e. no elements are comparable, so you only have to work out inclusions)? Maybe they are not even in bijection, I don't know.
$endgroup$
– Patrick Da Silva
Jul 19 '14 at 18:35
$begingroup$
@PatrickDaSilva $U$ is a set. It does not "have poset structure".
$endgroup$
– porton
Jul 19 '14 at 18:39
$begingroup$
Yes, sorry. I know what you mean, I was just thinking of a more general case (where $mathscr P(U)$ is replaced by a finite poset... but when I think about it now I don't know why I was thinking that). Still, have you checked the finite case, just cardinality-wise?
$endgroup$
– Patrick Da Silva
Jul 20 '14 at 7:00
$begingroup$
Actually I just checked it ; in the finite case, the order-isomorphism is essentially just ''adding decoration'' (i.e. mapping a filter with some set of generators to the filter which is generated by the filter with the same set of generators...). It's essentially because in this case $mathscr P(U) simeq F(mathscr P(U))$ so the question becomes boring. Interesting question.
$endgroup$
– Patrick Da Silva
Jul 20 '14 at 8:02
add a comment |
$begingroup$
Let $F(X)$ be the set of all filters (including the improper filter) on a poset $X$, ordered reversely to set-theoretic inclusion of filters.
Let $U$ be a set. Is $F(F(mathscr{P}U))$ order isomorphic to $F(mathscr{P}U)$?
set-theory filters
asked Jul 19 '14 at 18:03
portonporton
1,92111229
$endgroup$
Let $F(X)$ be the set of all filters (including the improper filter) on a poset $X$, ordered reversely to set-theoretic inclusion of filters.
Let $U$ be a set. Is $F(F(mathscr{P}U))$ order isomorphic to $F(mathscr{P}U)$?
set-theory filters
set-theory filters
asked Jul 19 '14 at 18:03
portonporton
1,92111229
asked Jul 19 '14 at 18:03
portonporton
1,92111229
edited Jul 19 '14 at 18:16
porton
asked Jul 19 '14 at 18:03
portonporton
1,92111229
asked Jul 19 '14 at 18:03
portonporton
1,92111229
asked Jul 19 '14 at 18:03
portonporton
1,92111229
1,92111229
$begingroup$
Have you checked the finite case where $U$ has a trivial poset structure (i.e. no elements are comparable, so you only have to work out inclusions)? Maybe they are not even in bijection, I don't know.
$endgroup$
– Patrick Da Silva
Jul 19 '14 at 18:35
$begingroup$
@PatrickDaSilva $U$ is a set. It does not "have poset structure".
$endgroup$
– porton
Jul 19 '14 at 18:39
$begingroup$
Yes, sorry. I know what you mean, I was just thinking of a more general case (where $mathscr P(U)$ is replaced by a finite poset... but when I think about it now I don't know why I was thinking that). Still, have you checked the finite case, just cardinality-wise?
$endgroup$
– Patrick Da Silva
Jul 20 '14 at 7:00
$begingroup$
Actually I just checked it ; in the finite case, the order-isomorphism is essentially just ''adding decoration'' (i.e. mapping a filter with some set of generators to the filter which is generated by the filter with the same set of generators...). It's essentially because in this case $mathscr P(U) simeq F(mathscr P(U))$ so the question becomes boring. Interesting question.
$endgroup$
– Patrick Da Silva
Jul 20 '14 at 8:02
add a comment |
$begingroup$
Have you checked the finite case where $U$ has a trivial poset structure (i.e. no elements are comparable, so you only have to work out inclusions)? Maybe they are not even in bijection, I don't know.
$endgroup$
– Patrick Da Silva
Jul 19 '14 at 18:35
$begingroup$
@PatrickDaSilva $U$ is a set. It does not "have poset structure".
$endgroup$
– porton
Jul 19 '14 at 18:39
$begingroup$
Yes, sorry. I know what you mean, I was just thinking of a more general case (where $mathscr P(U)$ is replaced by a finite poset... but when I think about it now I don't know why I was thinking that). Still, have you checked the finite case, just cardinality-wise?
$endgroup$
– Patrick Da Silva
Jul 20 '14 at 7:00
$begingroup$
Actually I just checked it ; in the finite case, the order-isomorphism is essentially just ''adding decoration'' (i.e. mapping a filter with some set of generators to the filter which is generated by the filter with the same set of generators...). It's essentially because in this case $mathscr P(U) simeq F(mathscr P(U))$ so the question becomes boring. Interesting question.
$endgroup$
– Patrick Da Silva
Jul 20 '14 at 8:02
$begingroup$
Have you checked the finite case where $U$ has a trivial poset structure (i.e. no elements are comparable, so you only have to work out inclusions)? Maybe they are not even in bijection, I don't know.
$endgroup$
– Patrick Da Silva
Jul 19 '14 at 18:35
$begingroup$
Have you checked the finite case where $U$ has a trivial poset structure (i.e. no elements are comparable, so you only have to work out inclusions)? Maybe they are not even in bijection, I don't know.
$endgroup$
– Patrick Da Silva
Jul 19 '14 at 18:35
$begingroup$
@PatrickDaSilva $U$ is a set. It does not "have poset structure".
$endgroup$
– porton
Jul 19 '14 at 18:39
$begingroup$
@PatrickDaSilva $U$ is a set. It does not "have poset structure".
$endgroup$
– porton
Jul 19 '14 at 18:39
$begingroup$
Yes, sorry. I know what you mean, I was just thinking of a more general case (where $mathscr P(U)$ is replaced by a finite poset... but when I think about it now I don't know why I was thinking that). Still, have you checked the finite case, just cardinality-wise?
$endgroup$
– Patrick Da Silva
Jul 20 '14 at 7:00
$begingroup$
Yes, sorry. I know what you mean, I was just thinking of a more general case (where $mathscr P(U)$ is replaced by a finite poset... but when I think about it now I don't know why I was thinking that). Still, have you checked the finite case, just cardinality-wise?
$endgroup$
– Patrick Da Silva
Jul 20 '14 at 7:00
$begingroup$
Actually I just checked it ; in the finite case, the order-isomorphism is essentially just ''adding decoration'' (i.e. mapping a filter with some set of generators to the filter which is generated by the filter with the same set of generators...). It's essentially because in this case $mathscr P(U) simeq F(mathscr P(U))$ so the question becomes boring. Interesting question.
$endgroup$
– Patrick Da Silva
Jul 20 '14 at 8:02
$begingroup$
Actually I just checked it ; in the finite case, the order-isomorphism is essentially just ''adding decoration'' (i.e. mapping a filter with some set of generators to the filter which is generated by the filter with the same set of generators...). It's essentially because in this case $mathscr P(U) simeq F(mathscr P(U))$ so the question becomes boring. Interesting question.
$endgroup$
– Patrick Da Silva
Jul 20 '14 at 8:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, they are not isomorphic if $U$ is infinite. Note first that $F(mathscr{P}U)$ has the property that every element is the join of the atoms below it (i.e., every filter on $U$ is the intersection of the ultrafilters containing it).
On the other hand, I claim $F(F(mathscr{P}U))$ does not have this property. First, the atoms of $F(F(mathscr{P}U))$ are the maximal filters on $F(mathscr{P}U)$. Any maximal filter $M$ on $F(mathscr{P}U)$ is principal (just take the union of all its elements, which will again be a proper filter on $U$ and must be in $M$ by maximality), and thus the maximal filters are exactly the principal filters generated by ultrafilters on $U$. Given an ultrafiler $omega$ on $U$, let us write $M_omega$ for the corresponding maximal filter on $F(mathscr{P}U)$, i.e. atom in $F(F(mathscr{P}U))$. Note notice that the join in $F(F(mathscr{P}U))$ of a collection of atoms $M_{omega_i}$ is just the principal filter on $F(mathscr{P}U)$ generated by $bigcap_i omega_i$. So, any non-principal filter on $F(mathscr{P}U)$ is not a join of atoms.
Examples of non-principal filters on $F(mathscr{P}U)$ are easy to find using Stone duality, which identifies $F(mathscr{P}U)$ with the lattice of closed subsets of $beta U$ (the space of ultrafilters on $U$), ordered by inclusion. In particular, for instance, if $omegainbeta U$ is any non-isolated point (i.e., it is a non-principal ultrafilter on $U$), then you could take the filter of all of closed neighborhoods of $omega$. Or, you could take any strictly decreasing sequence of closed sets $C_1supset C_2supsetdots$ (such a sequence is easy to construct if $U$ is infinite) and take the filter generated by the $C_n$.
answered Dec 19 '18 at 6:26
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f871921%2ffilters-on-a-set-of-filters-are-they-equivalent-to-just-filters%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, they are not isomorphic if $U$ is infinite. Note first that $F(mathscr{P}U)$ has the property that every element is the join of the atoms below it (i.e., every filter on $U$ is the intersection of the ultrafilters containing it).
On the other hand, I claim $F(F(mathscr{P}U))$ does not have this property. First, the atoms of $F(F(mathscr{P}U))$ are the maximal filters on $F(mathscr{P}U)$. Any maximal filter $M$ on $F(mathscr{P}U)$ is principal (just take the union of all its elements, which will again be a proper filter on $U$ and must be in $M$ by maximality), and thus the maximal filters are exactly the principal filters generated by ultrafilters on $U$. Given an ultrafiler $omega$ on $U$, let us write $M_omega$ for the corresponding maximal filter on $F(mathscr{P}U)$, i.e. atom in $F(F(mathscr{P}U))$. Note notice that the join in $F(F(mathscr{P}U))$ of a collection of atoms $M_{omega_i}$ is just the principal filter on $F(mathscr{P}U)$ generated by $bigcap_i omega_i$. So, any non-principal filter on $F(mathscr{P}U)$ is not a join of atoms.
Examples of non-principal filters on $F(mathscr{P}U)$ are easy to find using Stone duality, which identifies $F(mathscr{P}U)$ with the lattice of closed subsets of $beta U$ (the space of ultrafilters on $U$), ordered by inclusion. In particular, for instance, if $omegainbeta U$ is any non-isolated point (i.e., it is a non-principal ultrafilter on $U$), then you could take the filter of all of closed neighborhoods of $omega$. Or, you could take any strictly decreasing sequence of closed sets $C_1supset C_2supsetdots$ (such a sequence is easy to construct if $U$ is infinite) and take the filter generated by the $C_n$.
answered Dec 19 '18 at 6:26
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
No, they are not isomorphic if $U$ is infinite. Note first that $F(mathscr{P}U)$ has the property that every element is the join of the atoms below it (i.e., every filter on $U$ is the intersection of the ultrafilters containing it).
On the other hand, I claim $F(F(mathscr{P}U))$ does not have this property. First, the atoms of $F(F(mathscr{P}U))$ are the maximal filters on $F(mathscr{P}U)$. Any maximal filter $M$ on $F(mathscr{P}U)$ is principal (just take the union of all its elements, which will again be a proper filter on $U$ and must be in $M$ by maximality), and thus the maximal filters are exactly the principal filters generated by ultrafilters on $U$. Given an ultrafiler $omega$ on $U$, let us write $M_omega$ for the corresponding maximal filter on $F(mathscr{P}U)$, i.e. atom in $F(F(mathscr{P}U))$. Note notice that the join in $F(F(mathscr{P}U))$ of a collection of atoms $M_{omega_i}$ is just the principal filter on $F(mathscr{P}U)$ generated by $bigcap_i omega_i$. So, any non-principal filter on $F(mathscr{P}U)$ is not a join of atoms.
Examples of non-principal filters on $F(mathscr{P}U)$ are easy to find using Stone duality, which identifies $F(mathscr{P}U)$ with the lattice of closed subsets of $beta U$ (the space of ultrafilters on $U$), ordered by inclusion. In particular, for instance, if $omegainbeta U$ is any non-isolated point (i.e., it is a non-principal ultrafilter on $U$), then you could take the filter of all of closed neighborhoods of $omega$. Or, you could take any strictly decreasing sequence of closed sets $C_1supset C_2supsetdots$ (such a sequence is easy to construct if $U$ is infinite) and take the filter generated by the $C_n$.
answered Dec 19 '18 at 6:26
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
No, they are not isomorphic if $U$ is infinite. Note first that $F(mathscr{P}U)$ has the property that every element is the join of the atoms below it (i.e., every filter on $U$ is the intersection of the ultrafilters containing it).
On the other hand, I claim $F(F(mathscr{P}U))$ does not have this property. First, the atoms of $F(F(mathscr{P}U))$ are the maximal filters on $F(mathscr{P}U)$. Any maximal filter $M$ on $F(mathscr{P}U)$ is principal (just take the union of all its elements, which will again be a proper filter on $U$ and must be in $M$ by maximality), and thus the maximal filters are exactly the principal filters generated by ultrafilters on $U$. Given an ultrafiler $omega$ on $U$, let us write $M_omega$ for the corresponding maximal filter on $F(mathscr{P}U)$, i.e. atom in $F(F(mathscr{P}U))$. Note notice that the join in $F(F(mathscr{P}U))$ of a collection of atoms $M_{omega_i}$ is just the principal filter on $F(mathscr{P}U)$ generated by $bigcap_i omega_i$. So, any non-principal filter on $F(mathscr{P}U)$ is not a join of atoms.
Examples of non-principal filters on $F(mathscr{P}U)$ are easy to find using Stone duality, which identifies $F(mathscr{P}U)$ with the lattice of closed subsets of $beta U$ (the space of ultrafilters on $U$), ordered by inclusion. In particular, for instance, if $omegainbeta U$ is any non-isolated point (i.e., it is a non-principal ultrafilter on $U$), then you could take the filter of all of closed neighborhoods of $omega$. Or, you could take any strictly decreasing sequence of closed sets $C_1supset C_2supsetdots$ (such a sequence is easy to construct if $U$ is infinite) and take the filter generated by the $C_n$.
answered Dec 19 '18 at 6:26
Eric WofseyEric Wofsey
187k14215344
$endgroup$
No, they are not isomorphic if $U$ is infinite. Note first that $F(mathscr{P}U)$ has the property that every element is the join of the atoms below it (i.e., every filter on $U$ is the intersection of the ultrafilters containing it).
On the other hand, I claim $F(F(mathscr{P}U))$ does not have this property. First, the atoms of $F(F(mathscr{P}U))$ are the maximal filters on $F(mathscr{P}U)$. Any maximal filter $M$ on $F(mathscr{P}U)$ is principal (just take the union of all its elements, which will again be a proper filter on $U$ and must be in $M$ by maximality), and thus the maximal filters are exactly the principal filters generated by ultrafilters on $U$. Given an ultrafiler $omega$ on $U$, let us write $M_omega$ for the corresponding maximal filter on $F(mathscr{P}U)$, i.e. atom in $F(F(mathscr{P}U))$. Note notice that the join in $F(F(mathscr{P}U))$ of a collection of atoms $M_{omega_i}$ is just the principal filter on $F(mathscr{P}U)$ generated by $bigcap_i omega_i$. So, any non-principal filter on $F(mathscr{P}U)$ is not a join of atoms.
Examples of non-principal filters on $F(mathscr{P}U)$ are easy to find using Stone duality, which identifies $F(mathscr{P}U)$ with the lattice of closed subsets of $beta U$ (the space of ultrafilters on $U$), ordered by inclusion. In particular, for instance, if $omegainbeta U$ is any non-isolated point (i.e., it is a non-principal ultrafilter on $U$), then you could take the filter of all of closed neighborhoods of $omega$. Or, you could take any strictly decreasing sequence of closed sets $C_1supset C_2supsetdots$ (such a sequence is easy to construct if $U$ is infinite) and take the filter generated by the $C_n$.
answered Dec 19 '18 at 6:26
Eric WofseyEric Wofsey
187k14215344
answered Dec 19 '18 at 6:26
Eric WofseyEric Wofsey
187k14215344
answered Dec 19 '18 at 6:26
Eric WofseyEric Wofsey
187k14215344
answered Dec 19 '18 at 6:26
Eric WofseyEric Wofsey
187k14215344
187k14215344
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f871921%2ffilters-on-a-set-of-filters-are-they-equivalent-to-just-filters%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Have you checked the finite case where $U$ has a trivial poset structure (i.e. no elements are comparable, so you only have to work out inclusions)? Maybe they are not even in bijection, I don't know.
$endgroup$
– Patrick Da Silva
Jul 19 '14 at 18:35
$begingroup$
@PatrickDaSilva $U$ is a set. It does not "have poset structure".
$endgroup$
– porton
Jul 19 '14 at 18:39
$begingroup$
Yes, sorry. I know what you mean, I was just thinking of a more general case (where $mathscr P(U)$ is replaced by a finite poset... but when I think about it now I don't know why I was thinking that). Still, have you checked the finite case, just cardinality-wise?
$endgroup$
– Patrick Da Silva
Jul 20 '14 at 7:00
$begingroup$
Actually I just checked it ; in the finite case, the order-isomorphism is essentially just ''adding decoration'' (i.e. mapping a filter with some set of generators to the filter which is generated by the filter with the same set of generators...). It's essentially because in this case $mathscr P(U) simeq F(mathscr P(U))$ so the question becomes boring. Interesting question.
$endgroup$
– Patrick Da Silva
Jul 20 '14 at 8:02